Finding a quadratic equation given range and vertex to origin distance"Find the equation of quadratic function that value positive for −7&lt;x&lt;1 and the distance of vertex and origin is 5"

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Finding a quadratic equation given range and vertex to origin distance&quot;Find the equation of quadratic function that value positive for −7&amp;lt;x&amp;lt;1 and the distance of vertex and origin is 5&quot;

gil001q4wq · Accepted Answer

So a quadratic with roots at 1 and -7 will be c(x+7)(x−1), where c is a real number. We just have to figure out what c should be. We need c to be negative if you want the function to be positive on only (-7,1). Also, here the vertex is the point (-db2a,f(-db2a)). Let&#039;s write -t for our number c (so t will be positive), so our polynomial will be:f(x)=c(x2+6x−7)=cx2+6cx−7cTo find what c is, we assume that the distance between the vertex and the origin is 5 as desired. Compute the vertex of f(x):−b2a=−6c2c=−3f(−b2a)=f(−3)=c((−3)2+6(−3)−7)=c(9−18−7)=−16cwhich is what it means for the vertex to be:v=(−b2a,f(−b2a))=(−3,−16c)The distance between the vertex and the origin being 5 means that:5=(−3−0)2+(−16c−0)2=9+256c2solving for c gives c=d-14 (there are two solutions but we need the negative one like I stated).

enchantsyseq · Accepted Answer

You can use vertex form of the quadratic function which is given by,y−k=a(x−h)2.Given y is positive for −7&amp;lt;x&amp;lt;1, we take these points where the function y turns from positive to negative.Equating y to zero, we have x=h±−kaSolving, h−−ka=−7 and h+−ka=1h=−3,−ka=16As h2+k2=25,k=±4 We know the function is positive for x∈(−7,1) so k=4.a=−14The equation of quadratic is y=−14(x+3)2+4

Finding a quadratic equation given range and vertex

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